Find $\int \dfrac{1}{x^2-6x+13}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12 \text{arcsin}\left(\dfrac{x-3}{2}\right)+C$ (Choice B) B $\dfrac12 \text{arctan}\left(\dfrac{x-3}{2}\right)+C$ (Choice C) C $\dfrac14 \text{arcsin}\left(\dfrac{x-3}{2}\right)+C$ (Choice D) D $\dfrac14 \text{arctan}\left(\dfrac{x-3}{2}\right)+C$
The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as $(x+ h)^2+ k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as $(x+ h)^2+ k^2$ : $\begin{aligned} x^2-6x+13&=x^2-6x+9+4 \\\\ &=(x-3)^2+4 \\\\ &=(x{-3})^2+{2}^2 \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{x^2-6x+13}\,dx \\\\ &=\int\dfrac{1}{(x{-3})^2+{2}^2}\,dx \\\\ &=\dfrac{1}{{2}} \text{arctan}\left(\dfrac{x{-3}}{{2}}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{x^2-6x+13}\,dx=\dfrac12 \text{arctan}\left(\dfrac{x-3}{2}\right)+C$